Bảo toàn Hidro: \(n_{HCl}=2n_{H_2}=2\cdot\dfrac{3,36}{22,4}=0,3\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
\(\Rightarrow\) Chọn A
A. 0,6M
\(Fe + 2HCl \rightarrow FeCl_2 + H_2\)
\(n_{H_2} = \dfrac{3,36}{22,4}= 0,15 mol\)
Theo PTHH
\(n_{HCl} = 2n_{H_2}= 2 . 0,15 = 0,3 mol\)
CM\(HCl\)=\(\dfrac{0,3}{0,5}= 0,6M\)